Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int x;
    cin>>x;
    vector<double>dp(x+1);
    dp[0]=1.00;
    for(int i=1;i<=x;i++){
        double p;
        cin>>p;
        for(int j=i;j>=0;j--){
            
            dp[j]=((j!=0)?dp[j-1]*p:0)+dp[j]*(1-p);
        }
    }
    double ans=0;
    for(int i=1;i<=x;i++){
        // cout<<dp[i];
        if(x/2<i) ans+=dp[i];
    }
    printf("%.10lf",ans);
    return 0;
    
}